Question

x²+y²+z²=64,xy+yz+zx=18...find x+y+z

Answer 1

$Utilise \: the \: identity \: - \\ \\ (x + y + z {)}^{2} = {x}^{2} + y {}^{2} + {z}^{2} + 2(xy \: + \: yz \: + \: zx) \: \\ \\ Now \: , \: we're \: given \: that \: , \\ \\ {x}^{2} + y {}^{2} + z {}^{2} = 64 \\ xy+yz+zx= \: 18 \\ \\ Putting \: the \: values \: in \: the \: above \: \\ identity \: , \\ \\ (x + y + z {)}^{2} = 64 + 2(18) = 100 \\ \\ || \: \: (x + y + z) \: = \: 10 \: or - 10 \: \: ||$